3. | -9 | = 9
7. - | -5 * -7 | = - | 35 | = -35
15. | m + 3 | = 7
m + 3 = 7 -(m + 3) = 7
m = 7 - 3 -m - 3 = 7
m = 4 -m = 10
m = -10
so m = 4 and m = -10
17. | -3d | = 15
-3d = 15 -(-3d) = 15
d = 15/-3 3d = 15
d = - 5 d = 15/3
d = 5
so d = 5 and d = -5
19. | 4b - 5 | = 19
4b - 5 = 19 -(4b - 5) = 19
4b = 19 + 5 -4b + 5 = 19
4b = 24 -4b = 14
b = 24/4 b = 14/-4
b = 6 b = - 7/2
so b = 6 or b = -7/2
20. | x - 1 | + 5 = 2
| x - 1 | = 2 - 5
| x - 1 | = -3
There is no solution on this one because an absolute value cannot equal a negative number
21. -4 | 8 - 5n | = 13
| 8 - 5n | = - 13/4
There is also no solution for this one because of the same reason...absolute values cannot equal negative numbers
Answer:
5.4 units
Step-by-step explanation:
From the figure ,it's clear that sin(50) = PQ/7
Then
PQ = 7×sin(50) = 5.362311101833
Therefore the approximate length of PQ I saw 5.4
D, hope this helps because 1/2 is less than 1. and 5*1/2 is 2.5 and that is less than 5
Answer:
Step-by-step explanation:
We do not consider zero to be a positive or negative number. For each positive integer, there is a negative integer, and these integers are called opposites. For example, -3 is the opposite of 3, -21 is the opposite of 21, and 8 is the opposite of -8. ... If an integer is less than zero, we say that its sign is negative
The objective function is simply a function that is meant to be maximized. Because this function is multivariable, we know that with the applied constraints, the value that maximizes this function must be on the boundary of the domain described by these constraints. If you view the attached image, the grey section highlighted section is the area on the domain of the function which meets all defined constraints. (It is all of the inequalities plotted over one another). Your job would thus be to determine which value on the boundary maximizes the value of the objective function. In this case, since any contribution from y reduces the value of the objective function, you will want to make this value as low as possible, and make x as high as possible. Within the boundaries of the constraints, this thus maximizes the function at x = 5, y = 0.
